Go back N ARQ MCQ in Data Link Layer

1. In Go back N ARQ Duplicate packet problem can be solved by

A. Increasing receiving window size
B. Increasing sending window size
C. Increasing sequence number
D. Decreasing sequence number
E. None of the above

Ans- C. Increasing sequence number

Solution:- Duplicate packet problem can be solved by two ways:-

Ws+Wr=< Available Sequence Number

Increasing the size of sequence number
Decreasing the sending window size

2. In Go back N ARQ if sending window size is 3 and sender wants to send 8 packet. If every 4^th packet lost than How many transmission is required to send all these packets.

A.19
B. 18
C. 20
D. 22
E. None of the above

Ans- A.19

Total Number of transaction we get 19

Selective repeat MCQs

3. In Go Back N ARQ receiver window size is always…

A. 5
B. 6
C. 2
D. 1
E. None of the above

Ans- D. 1

3. In Go Back N ARQ receiver window size is always…

A. 5
B. 6
C. 2
D. 1
E. None of the above

Ans- D. 1

4. In Go Back 5 ARQ sender window size is…

A. 5
B. 6
C. 2
D. 1
E. None of the above

Ans- A. 5

5. In Go Back 5 ARQ available sequence number range is…

A. 0-6
B. 0-4
C. 0-5
D. 0-1
E. None of the above

Ans- C. 0-5

6. In Go Back N ARQ if sender window size is N than available sequence number range is…

A. 0- N -1
B. 0- N+1
C. 0 – N
D. 0-N/2
E. None of the above

Ans-C. 0 – N

Selective repeat MCQs

7. In Go Back N ARQ if highest sequence number is N than sender window size is …

A. N -1
B. N+1
C. N
D. N/2
E. None of the above

Ans-C. N

8. In Go Back N ARQ if size of sequence number is 6 bit than total sequence number is…

A. 62
B. 64
C. 66
D. 63
E. None of the above

Ans-B. 64

9. In Go Back N ARQ if size of sequence number is K bit than total sequence number is…

A. 2^k
B. 2^k-1
C. 2^k+1
D. 2^k+2
E. None of the above

Ans-A. 2^k

10. In Go Back N ARQ if size of sequence number is K bit than window sender size is…

A. 2^k
B. 2^k-1
C. 2^k+1
D. 2^k+2
E. None of the above

Ans-A. 2^k-1

11. In Go Back N ARQ if sender window size is 6 than min sequence number required is…

A. 8
B. 6
C. 5
D. 7
E. None of the above

Ans-D. 7

12. In Go Back N ARQ if 256 bit size packet is to be send on a link with 8 Kbps bandwidth. if one way propagation delay is 10msec and value of n is 10 than find efficiency and data rate of this link.

A. 8 and 64 Kbps
B. 8 and 128 Kbps
C. 16 and 256 Kbps
D. 24 and 32 Kbps
E. None of the above

Ans-A. 8 and 64 Kbps

Solution:-In Go back n ARQ Efficiency = N* Useful Time/Total Time

Useful Time= Transmission time
Total Time = Transmission time + Propagation time

Transmission time T_T=256/8*103=> 32 msec
Efficiency = N*T_T/T_T+T_p =>10*32/40=> 8

Throughput = Length of Frame * N /Total Time(Round Trip Time)

Total Time = Transmission time + Propagation time or Total time = Round trip time(RTT)
Throughput = Length of Frame * N / T_T+2T_p
now first we calculate Transmission time which is length of frame/link bandwidth

T_T=L/B_w
T_T=256/8*10³=> 32 msec
Now put the value of transmission time into Throughput formula

Throughput = Length of Frame * N / T_T+2T_p
Throughput = 10*256*10³/32+8 =>64 Kbps